Prove every odd integer is the difference of two squares

I know that I should use the definition of an odd integer ($2k+1$), but that's about it.

Thanks in advance!

Solution 1:

Step 1: pick an odd number (like $n=13$ here)

13 squares

Step 2: bend it in "half" (any odd number $n$ can be written as $2k+1$, and $13=2\cdot 6 + 1$)

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Step 3: fill in the blank space

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Step 4: Count squares. (Here, the blue square has area $36=6^2$, while the whole square has area $49=7^2$)

Hint: Consider the difference of two consecutive squares. What is $(k+1)^2-k^2$?

HINT: $$\begin{align} &2k + 1 \\= & 1\cdot(2k + 1) \\ =& \left(k + 1 - k \right)\left(k + 1 + k\right) \\ = & \cdots\end{align}$$